Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define each(itr,c) for(__typeof(c.begin()) itr=c.begin(); itr!=c.end(); ++itr)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second

typedef pair<int,int> pi;

const int N=100000, M=300000;
const int INF=12345678;

int d[2][N];
vector<int> G[N];

int d_ct=0;
void dijkstra(int s)
{
    // dijkstra
    priority_queue<pi,vector<pi>,greater<pi>> que;
    fill(d[d_ct],d[d_ct]+N,INF);
    d[d_ct][s]=0;
    que.push(pi(0,s));
    while(!que.empty())
    {
        pi p=que.top();
        que.pop();
        int v=p.se;
        if(d[d_ct][v]<p.fi) continue;
        rep(i,G[v].size())
        {
            int nx=G[v][i];
            if(d[d_ct][nx]>d[d_ct][v]+1)
            {
                d[d_ct][nx]=d[d_ct][v]+1;
                que.push(pi(d[d_ct][nx],nx));
            }
        }
    }
    ++d_ct;
}

int cts[N]={0}, ctt[N]={0};

int main()
{
    int n,m,s,t;
    scanf(" %d %d %d %d", &n, &m, &s, &t);
    --s;
    --t;
    rep(i,m)
    {
        int x,y;
        scanf(" %d %d", &x, &y);
        --x;
        --y;
        G[x].pb(y);
        G[y].pb(x);
    }

    dijkstra(s);
    dijkstra(t);

    rep(i,n) if(d[0][i]<N) ++cts[d[0][i]];
    rep(i,n) if(d[1][i]<N) ++ctt[d[1][i]];

    ll ans=0;
    rep(i,d[0][t]-1) ans+=(ll)cts[i]*ctt[d[0][t]-2-i];
    cout << ans << endl;
    return 0;
}